1.\(C_M=\dfrac{10.D.C\%}{M}=\dfrac{10.1,19.37}{36,5}\approx12,063M\)
2.\(C\%=\dfrac{C_M.M}{10.D}=\dfrac{10,81.36,5}{10.1,17}\approx33,724\%\)
Bài 1: Coi dd HCl đó chứa \(m=1,19g;V=1ml=0,001l\) (Phù hợp với D)
Theo bài ra ta có:
\(m_{HCl}=1,19.37\%=0,4403\left(g\right)\\ \Rightarrow n_{HCl}=\dfrac{0,4403}{36,5}=\dfrac{4403}{365000}\left(mol\right)\)
Do đó \(C_{M_{\text{dd}HCl}}=\dfrac{\dfrac{4403}{365000}}{0,001}\approx12,063M\)
Bài 2:
Coi dd HCl đó có
\(m=1,17g;V=1ml=0,001l\)
Theo gt ta có:
\(n_{HCl}=10,81.0,001=0,01081\left(mol\right)\\ \Rightarrow m_{HCl}=0,01081.36,5=0,394565\left(g\right)\)
Do đó \(\%C_{HCl}=\dfrac{0,394565.100\%}{1,17}\approx33,724\%4\%\)
